Question

The percentage volume of an ice block ( Relative Density = 0.96 ) above sea
water surface if it floats in that water (RD = 1.025 ) is
​

Answer 1

Answer:

0.063%

Explanation:

Percentage Volume

=( 1-(Vi/Vw))×100

=1-(0.96/1.025)×100

0.063%

Answer 2

Answer:

The percentage volume of an ice block above sea water surface is 6.4.

Explanation:

By the principle of floatation,

$\frac{v}{V} =\frac{D_i}{D_s}$        (1)

Where,

v=volume of iceberg immersed in water

V=total volume of iceberg

Di=relative density of ice

Ds=relative density of sea water

From the question we have,

Di=0.96

Ds=1.025

By substituting the value of Di and Ds in equation (1) we get;

$\frac{v}{V} =\frac{0.96}{1.025}$

$\frac{v}{V} =0.936$    (2)

The percentage volume of an ice block above sea water surface is,

$(1-\frac{v}{V})\times 100$       (3)

By substituting the values in equation (3) we get;

$(1-0.936)\times 100$

$0.064\times 100=6.4\%$

Hence, the percentage volume of an ice block above sea water surface is 6.4.